Alamofire: FAILURE responseSerializationFailed

血红的双手。 提交于 2021-01-29 05:57:35

问题


I'm trying to get the JSON from a URL to download using Alamofire.

Alamofire.request(requrl, method: .get, encoding: JSONEncoding.default)
        .responseJSON { response in
            print(response.result)
            print(response.value)
            debugPrint(response)


            }

However the value of response.value is nil, and the request result status is FAILURE. The following is in the logs:

>[Data]: 919 bytes
[Result]: FAILURE: responseSerializationFailed(Alamofire.AFError.ResponseSerializationFailureReason.jsonSerializationFailed(Error Domain=NSCocoaErrorDomain Code=3840 "Invalid value around character 0." UserInfo={NSDebugDescription=Invalid value around character 0.}))
[Timeline]: Timeline: { "Request Start Time": 561988246.258, "Initial Response Time": 561988246.446, "Request Completed Time": 561988246.447, "Serialization Completed Time": 561988246.447, "Latency": 0.189 secs, "Request Duration": 0.189 secs, "Serialization Duration": 0.000 secs, "Total Duration": 0.190 secs

The JSON url in a browser returns:

php script

<?php
header('Content-type: application/json');
//header('Content-Type: html; charset=utf-8');
$sqlstatement = $_GET["sqlstatement"];
// Create connection
$dbConnection=mysqli_connect("***.***.gear.host","****","*****!","****");

//$sqlstatement = "SELECT * FROM reeds.tbl_user";
// Check connection
if (mysqli_connect_errno())
{
  // Print error message 
  echo mysqli_connect_error();
}

// Stores SQL statement, selecting all objects from testcomputing.name


// Check to ensure results > = 1
if ($result = mysqli_query($dbConnection, $sqlstatement))
{
    // If so, then create a results array and a temporary one to hold the data
    $resultArray = array();
    $tempArray = array();

    // Loop through each row in the result set
    while($row = $result->fetch_object())
    {
        // Add each row into our results array
    $tempArray = $row;
     array_push($resultArray, $tempArray);
    }

    // Encode the array to JSON and output the results
    echo json_encode($resultArray);
}

// Close connections
mysqli_close($dbConnection);
?>

回答1:


The issue was associated with the hosting provider. The provider I was originally using had strange security protocols installed, as such no JSON was being returned, after switching it work perfectly.



来源:https://stackoverflow.com/questions/52948481/alamofire-failure-responseserializationfailed

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!